Potential energy can be calculated by the formula Pe=mgh. Plug in your values:
Pe=mgh
Pe=(6 kg)(9.8m/s^2)(100 m)
Pe=5880 kg x m^2/s^2, or 5880 Joules
Answer:
975.56×10²³ molecules
Explanation:
Given data:
Number of molecules of C₂H₆ = 4.88×10²⁵
Number of molecules of CO₂ produced = ?
Solution:
Chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Number of moles of C₂H₆:
1 mole = 6.022×10²³ molecules
4.88×10²⁵ molecules×1mol/6.022×10²³ molecules
0.81×10² mol
81 mol
Now we will compare the moles of C₂H₆ with CO₂.
C₂H₆ : CO₂
2 : 4
81 : 4/2×81 = 162 mol
Number of molecules of CO₂:
1 mole = 6.022×10²³ molecules
162 mol ×6.022×10²³ molecules / 1 mol
975.56×10²³ molecules
Answer:
The correct answer is 146 g/mol
Explanation:
<em>Freezing point depression</em> is a colligative property related to the number of particles of solute dissolved in a solvent. It is given by:
ΔTf = Kf x m
Where ΔTf is the freezing point depression (in ºC), Kf is a constant for the solvent and m is the molality of solution. From the problem, we know the following data:
ΔTf = 1.02ºC
Kf = 5.12ºC/m
From this, we can calculate the molality:
m = ΔTf/Kf = 1.02ºC/(5.12ºC/m)= 0.199 m
The molality of a solution is defined as the moles of solute per kg of solvent. Thus, we can multiply the molality by the mass of solvent in kg (250 g= 0.25 kg) to obtain the moles of solute:
0.199 mol/kg benzene x 0.25 kg = 0.0498 moles solute
There are 0.0498 moles of solute dissolved in the solution. To calculate the molar mass of the solute, we divide the mass (7.27 g) into the moles:
molar mass = mass/mol = 7.27 g/(0.0498 mol) = 145.9 g/mol ≅ 146 g/mol
<em>Therefore, the molar mass of the compound is 146 g/mol </em>
Answer:
Potassium
1s2 2s2 2p6 3s2 3p6 4s1
Explanation:
The atom having only one electron its outermost shell must belong to an element in group one of the periodic table.
Having noted that, we proceed to find out what element in group one that has the atom just described in the question.
That atom must belong to an element in the fourth period. The only group 1 element in the fourth period is potassium.
The electron configuration of potassium is;
1s2 2s2 2p6 3s2 3p6 4s1
